What is the integral of (8x^2+7x+8)/((x^2+1)^2) ?

The integral of (8x^2+7x+8)/((x^2+1)^2) is 8arctan(x)-7/(2(x^2+1))+C

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integral of (8x^2+7x+8)/((x^2+1)^2)

\int \frac{8 x ^{2} + 7 x + 8}{( x ^{2} + 1 ) ^{2}} d x

8 arctan (x) - \frac{7}{2 ( x ^{2} + 1 )} + C

Solution steps

\int \frac{8 x ^{2} + 7 x + 8}{( x ^{2} + 1 ) ^{2}} d x

Take the partial fraction of

\frac{8 x ^{2} + 7 x + 8}{( x ^{2} + 1 ) ^{2}} : \frac{8}{x ^{2} + 1} + \frac{7 x}{( x ^{2} + 1 ) ^{2}}

= \int \frac{8}{x ^{2} + 1} + \frac{7 x}{( x ^{2} + 1 ) ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{8}{x ^{2} + 1} d x + \int \frac{7 x}{( x ^{2} + 1 ) ^{2}} d x

\int \frac{8}{x ^{2} + 1} d x = 8 arctan (x)

\int \frac{7 x}{( x ^{2} + 1 ) ^{2}} d x = - \frac{7}{2 ( x ^{2} + 1 )}

= 8 arctan (x) - \frac{7}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= 8 arctan (x) - \frac{7}{2 ( x ^{2} + 1 )} + C

x \to 5 lim (f (x))

f (x) = lo g_{4} (x^{2} - 3 x)

\frac{d y}{d x} = e^{5 y + 6 x}, y (0) = 2

\frac{d}{d x} (\frac{( 1 - 5 x )}{( 1 + 5 x )})

\frac{d y}{d x} = y (1 - 0.0002 y)

What is the integral of (8x^2+7x+8)/((x^2+1)^2) ?
The integral of (8x^2+7x+8)/((x^2+1)^2) is 8arctan(x)-7/(2(x^2+1))+C